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How do you figure out the length of two sides of a right triangle knowing the angles + the length of one side?

How do you figure out the length of two sides of a right triangle knowing the angles + the length of one side? Topic: How do you figure out the length of two sides of a right triangle knowing the angles + the length of one side?
July 20, 2019 / By Debi
Question: I have a problem with a right triangle, I know how long the base is and all the interior angles. How do I figure out the length of the other two sides? Find X and Y Right triangle with a base of 14, X is the hight, Y is the diagonal, and the angle between the base and Y is 30 degrees. I am getting kinda confused here.
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Best Answers: How do you figure out the length of two sides of a right triangle knowing the angles + the length of one side?

Britt Britt | 1 day ago
law of sines uppercase is angle measurement and lowercase is the side that is opposite of that angle (sinA)/a = (sinB)/b = (sinC)/c solve the proportions to find the length of the sides
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Britt Originally Answered: How can I solve for an unknown side of a right triangle if I am only given one side length?
Hello, I am sorry but you need three independent data to define univocally a triangle. Here you only provide two of them: ● We are talking of a right triangle. ● One side of the triangle is 6. Hence, except you provide another fact, there is an infinity of triangles that would fit your description. = = = = = = = = = = = = = Hello again, It seems your problem have a drawing that is essential to its solving. The best thing would be to provide the full exercise (complete with drawing). Take a picture of it, upload it into ImageShack and paste the link here. We'll see then. Logically, Dragon.Jade :-)

Aline Aline
advantageous. right here is going. permit's think of a triangle with the wonderful perspective backside left, short side vertical, horizontal side slightly longer and going to the wonderful. Draw one for your self. permit's call the wonderful perspective A, the dazzling left vertex B, and the only on the backside right C. permit's say that all of us understand perspective B and this is 30º. permit's say that all of us understand side AB and this is 5. (use any values you like, those are merely so as that we are able to communicate approximately a thank you to do it). all of us understand that the adjoining side to the perspective, side AB is a ingredient of the equation cos 30º = 5/BC. We remedy for the hypotenuse, BC, to get BC = 5 / cos30º or approximately 5.seventy seven (this is precisely 10/sqrt3). Now you have the hypotenuse and an ingredient. you ought to use the Pythagorean theorem to get 2.89 for the different side, or set up yet another trig equation. (that's going to be opposite the perspective you recognize, so which you will possibly use sine). desire this facilitates. Cheers.
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Ulick Ulick
if u consider ur base is opposite side to the angle u know than simply sin theta = (opp side) / (hyp) as you know the theta , then you know the value of sin theta also and also u know the value of opp side that is ur base value. after substituing these values u get the hypotenuse value. as per pythagorous theorem, sqr(hyp) = sqr(opp side) + sqr (adj side) substitue the values of opp side and hyp in the above formula , you will get the value of adj side. so now you know length of all the sides of right angled triangle.. hope i'm clear in my explanation.
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Rimon Rimon
sohcahtoa sinx = opposite/hypotenuse cosx = adjacent/hypotenuse tanx = opposite/adjacent where hypotenuse is the side that is opposite of right angle opposite is the side that is opposite of the angle x and adjacent is the third side
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Rimon Originally Answered: Trig problem: triangle ABC. angle B=48, AC=13.6, BC=15.3. how do i find the other angles and sides?
If the triangle is not a right-triangle. You should use the law of Sines or Law of Cosines to help. I STRONGLY advise Law of Sines since you have 2 sides and 1 angle. The law of sines is 1 big proportion using sin. It looks like: (sinA/a)=(sinB/b)=(sinC/c) Good luck on your exam!

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